# Real and imaginary numbers?

1. Dec 10, 2008

### matt_crouch

Im doing A-level maths at the moment an its nowhere in the sylabus im just generally interested :D

can anyone giv me like a really simplish explaination :D
cheers

2. Dec 10, 2008

### mathwonk

real numbers are in 1-1 correspondence with points of a line. complex (i.e. "imaginary") numbers are in 1-1 correspondence with points of the plane, and include the real numbers as the special case of the points of the x axis in the plane.

thus a complex number has 2 coordinates (a,b). To add (a,b)+(x,y) = (a+x,b+y) you add the coordinates separately. to multiply (a,b).(x,y) is more complicated.

the rule is: (a,b).(x,y) = (ax-by, ay+bx). the real numbers are the ones of form (x,0), or just x. Thus notice that (0,1).(0,1) = (-1,0). So we now have a new non real number namely (0,1), whose square equals the real number -1.

If we give a special name, usually i, to the number (0,1), then we can write every complex number (a,b) as a(1,0) + b(0,1) = a + bi, where i^2 = -1.

This explains the multiplication rule, since now we see that (a,b).(x,y) = (a+bi)(x+iy)

= ab + xyi^2 +ayi + bxi = (ab-xy) + i(ay+bx) = (ab-xy, ay+bx).

If we also remember the addition laws for sin and cosine, we get that multiplication of two complex numbers, multiplies their lengths (distance from the origin) and adds the angles made by the lines joining them to the origin.

thus we can write every complex number as r(cos(t)+ i sin(t)),

and we have [rcos(t) + i sin(t)][scosu)+i sin(u)] = rs[cos(s+t) + i sin(s+t)]

hows them apples?

3. Dec 10, 2008

### arildno

As a proud Norwegian, I will add that a compatriot of mine, Caspar Wessel, was the first to formulate the idea of complex numbers as operations done in the "plane", rather than "on the line".

In his original paper, he conceives of the idea of how to "multiply" line segments (all initiating at the origin), whereby he arrived at the strategy of that the product should be a line segment whose length should be the product of the factors' length, and the product line segment's angle to the x-axis should be the sum of the angles the factors made to the x-axis.

Here is a biographical sketch of him, from MacTutor:
http://www-history.mcs.st-andrews.ac.uk/Biographies/Wessel.html

Last edited: Dec 10, 2008
4. Dec 10, 2008

### symbolipoint

A simpler explanation, certainly less thorough than Mathwonk's just given, is that the simplest imaginary number is the solution to the equation, x^2 + 1 = 0. The solution to this equation is the number called i.

Last edited: Dec 10, 2008
5. Dec 10, 2008

### lurflurf

Do we have a circular reasoning icon in this forum? How can we define a number as being a solution of an equation without having first defined the number? We need to be more formal, let C=R be a field, that contains R (real numbers) and an element i such that i^2=-1 and closed under field operations.
Theorem
All comblex numbers may be written in the form a+bi where a and b are real numbers.
Theorem
(a+bi)+(c+di)=(a+c)+(d+d)i

6. Jan 28, 2009

### Chewy0087

Sorry for the bump but didn't want to make a new thread for such an easy question >.<

i * -i ? Hmm :P, logic would maybe say 1 but maybe i'm wrong? Any clarification would be appreciated

7. Jan 28, 2009

### qntty

You are correct, i * -i = 1.
$i \cdot -i = -i^2 = -\sqrt{-1}^2 = -1 \cdot -1 = 1$

8. Jan 28, 2009

### NoMoreExams

i*-i = -i^2 = -1*(-1) = 1.